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Dacă a=1,(72) si b=3,1 (24) aflati perimetrul scenei scriind rezultatul sub forma de fractie ireductibila.

Răspuns :

1,(72)=172-1)/99=171/99=57/33=19/11
3,1(24)=3124-31)/990=3093/990=1031/330
P=2*19/11+2*1031/330
=38/11+2062/330
=1140+2062)/330
=3202/330
=1601/165
1,(72)=172-1)/99=171/99=57/33=19/11
3,1(24)=3124-31)/990=3093/990=1031/330
P=2(19/11+1031/330)=38/11+2062/330=(1140+2062)/330=3202/330=1601/165